文章目录
链表
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输入链表头节点，奇数长度返回中点，偶数长度返回上中点
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输入链表头节点，奇数长度返回中点，偶数长度返回下中点
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输入链表头节点，奇数长度返回中点前一个，偶数长度返回上中点前一个
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输入链表头节点，奇数长度返回中点前一个，偶数长度返回下中点前一个
public class LinkedListMid {
public static class Node {
public int value;
public Node next;
public Node(int v) {
value = v;
}
}
/**
* 输入链表头节点，奇数长度返回中点，偶数长度返回上中点
*/
public static Node midOrUpMidNode(Node head) {
//head代表链表，head.next代表第一个节点，head.next.next代表第二个节点
if (head == null || head.next == null || head.next.next == null) {
return head;
}
//链表有三个节点及以上
Node slow = head.next;//慢指针
Node fast = head.next.next;//快指针
/**
* 原理：慢指针走一步，快指针走两步。快指针走到终点的时候，慢指针刚好走到中间位置
*/
while (fast.next != null && fast.next.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
/**
* 输入链表头节点，奇数长度返回中点，偶数长度返回下中点
*/
public static Node midOrDownMidNode(Node head) {
if (head == null || head.next == null) {
return head;
}
Node slow = head.next;
Node fast = head.next;
while (fast.next != null && fast.next.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
/**
* 输入链表头节点，奇数长度返回中点前一个，偶数长度返回上中点前一个
*/
public static Node midOrUpMidPreNode(Node head) {
if (head == null || head.next == null || head.next.next == null) {
return null;
}
Node slow = head;
Node fast = head.next.next;
while (fast.next != null && fast.next.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
/**
* 输入链表头节点，奇数长度返回中点前一个，偶数长度返回下中点前一个
*/
public static Node midOrDownMidPreNode(Node head) {
if (head == null || head.next == null) {
return null;
}
if (head.next.next == null) {
return head;
}
Node slow = head;
Node fast = head.next;
while (fast.next != null && fast.next.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
/**
* 以下方法使用ArrayList做以上的操作，不使用链表
*/
public static Node right1(Node head) {
if (head == null) {
return null;
}
Node cur = head;
ArrayList<Node> arr = new ArrayList<>();
while (cur != null) {
arr.add(cur);
cur = cur.next;
}
return arr.get((arr.size() - 1) / 2);
}
public static Node right2(Node head) {
if (head == null) {
return null;
}
Node cur = head;
ArrayList<Node> arr = new ArrayList<>();
while (cur != null) {
arr.add(cur);
cur = cur.next;
}
return arr.get(arr.size() / 2);
}
public static Node right3(Node head) {
if (head == null || head.next == null || head.next.next == null) {
return null;
}
Node cur = head;
ArrayList<Node> arr = new ArrayList<>();
while (cur != null) {
arr.add(cur);
cur = cur.next;
}
return arr.get((arr.size() - 3) / 2);
}
public static Node right4(Node head) {
if (head == null || head.next == null) {
return null;
}
Node cur = head;
ArrayList<Node> arr = new ArrayList<>();
while (cur != null) {
arr.add(cur);
cur = cur.next;
}
return arr.get((arr.size() - 2) / 2);
}
public static void main(String[] args) {
Node test = null;
test = new Node(0);
test.next = new Node(1);
test.next.next = new Node(2);
test.next.next.next = new Node(3);
test.next.next.next.next = new Node(4);
test.next.next.next.next.next = new Node(5);
test.next.next.next.next.next.next = new Node(6);
test.next.next.next.next.next.next.next = new Node(7);
test.next.next.next.next.next.next.next.next = new Node(8);
Node ans1 = null;
Node ans2 = null;
ans1 = midOrUpMidNode(test);
ans2 = right1(test);
System.out.println(ans1 != null ? ans1.value : "无");
System.out.println(ans2 != null ? ans2.value : "无");
ans1 = midOrDownMidNode(test);
ans2 = right2(test);
System.out.println(ans1 != null ? ans1.value : "无");
System.out.println(ans2 != null ? ans2.value : "无");
ans1 = midOrUpMidPreNode(test);
ans2 = right3(test);
System.out.println(ans1 != null ? ans1.value : "无");
System.out.println(ans2 != null ? ans2.value : "无");
ans1 = midOrDownMidPreNode(test);
ans2 = right4(test);
System.out.println(ans1 != null ? ans1.value : "无");
System.out.println(ans2 != null ? ans2.value : "无");
}
}
/**
*如：12321，abcba等这样的属于回文结构
*/
public class IsPalindromeList {
public static class Node {
public int value;
public Node next;
public Node(int data) {
this.value = data;
}
}
// need n extra space
public static boolean isPalindrome1(Node head) {
Stack<Node> stack = new Stack<Node>();
Node cur = head;
while (cur != null) {
stack.push(cur);
cur = cur.next;
}
while (head != null) {
if (head.value != stack.pop().value) {
return false;
}
head = head.next;
}
return true;
}
// need n/2 extra space
public static boolean isPalindrome2(Node head) {
if (head == null || head.next == null) {
return true;
}
Node right = head.next;
Node cur = head;
while (cur.next != null && cur.next.next != null) {
right = right.next;
cur = cur.next.next;
}
Stack<Node> stack = new Stack<Node>();
while (right != null) {
stack.push(right);
right = right.next;
}
while (!stack.isEmpty()) {
if (head.value != stack.pop().value) {
return false;
}
head = head.next;
}
return true;
}
// need O(1) extra space
public static boolean isPalindrome3(Node head) {
if (head == null || head.next == null) {
return true;
}
Node n1 = head;
Node n2 = head;
while (n2.next != null && n2.next.next != null) { // find mid node
n1 = n1.next; // n1 -> mid
n2 = n2.next.next; // n2 -> end
}
n2 = n1.next; // n2 -> right part first node
n1.next = null; // mid.next -> null
Node n3 = null;
while (n2 != null) { // right part convert
n3 = n2.next; // n3 -> save next node
n2.next = n1; // next of right node convert
n1 = n2; // n1 move
n2 = n3; // n2 move
}
n3 = n1; // n3 -> save last node
n2 = head;// n2 -> left first node
boolean res = true;
while (n1 != null && n2 != null) { // check palindrome
if (n1.value != n2.value) {
res = false;
break;
}
n1 = n1.next; // left to mid
n2 = n2.next; // right to mid
}
n1 = n3.next;
n3.next = null;
while (n1 != null) { // recover list
n2 = n1.next;
n1.next = n3;
n3 = n1;
n1 = n2;
}
return res;
}
public static void printLinkedList(Node node) {
System.out.print("Linked List: ");
while (node != null) {
System.out.print(node.value + " ");
node = node.next;
}
System.out.println();
}
public static void main(String[] args) {
Node head = null;
printLinkedList(head);
System.out.print(isPalindrome1(head) + " | ");
System.out.print(isPalindrome2(head) + " | ");
System.out.println(isPalindrome3(head) + " | ");
printLinkedList(head);
System.out.println("=========================");
head = new Node(1);
printLinkedList(head);
System.out.print(isPalindrome1(head) + " | ");
System.out.print(isPalindrome2(head) + " | ");
System.out.println(isPalindrome3(head) + " | ");
printLinkedList(head);
System.out.println("=========================");
head = new Node(1);
head.next = new Node(2);
printLinkedList(head);
System.out.print(isPalindrome1(head) + " | ");
System.out.print(isPalindrome2(head) + " | ");
System.out.println(isPalindrome3(head) + " | ");
printLinkedList(head);
System.out.println("=========================");
head = new Node(1);
head.next = new Node(1);
printLinkedList(head);
System.out.print(isPalindrome1(head) + " | ");
System.out.print(isPalindrome2(head) + " | ");
System.out.println(isPalindrome3(head) + " | ");
printLinkedList(head);
System.out.println("=========================");
head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
printLinkedList(head);
System.out.print(isPalindrome1(head) + " | ");
System.out.print(isPalindrome2(head) + " | ");
System.out.println(isPalindrome3(head) + " | ");
printLinkedList(head);
System.out.println("=========================");
head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(1);
printLinkedList(head);
System.out.print(isPalindrome1(head) + " | ");
System.out.print(isPalindrome2(head) + " | ");
System.out.println(isPalindrome3(head) + " | ");
printLinkedList(head);
System.out.println("=========================");
head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(1);
printLinkedList(head);
System.out.print(isPalindrome1(head) + " | ");
System.out.print(isPalindrome2(head) + " | ");
System.out.println(isPalindrome3(head) + " | ");
printLinkedList(head);
System.out.println("=========================");
head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(2);
head.next.next.next = new Node(1);
printLinkedList(head);
System.out.print(isPalindrome1(head) + " | ");
System.out.print(isPalindrome2(head) + " | ");
System.out.println(isPalindrome3(head) + " | ");
printLinkedList(head);
System.out.println("=========================");
head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(2);
head.next.next.next.next = new Node(1);
printLinkedList(head);
System.out.print(isPalindrome1(head) + " | ");
System.out.print(isPalindrome2(head) + " | ");
System.out.println(isPalindrome3(head) + " | ");
printLinkedList(head);
System.out.println("=========================");
}
}
/**
* 将单向链表按某值划分成左边小，中间相等，右边大的形式
*/
public class SmallerEqualBigger {
/**
* 方法1：将链表放入数组中，在数组上做partition（时间复杂度很优了O（n），只不过使用了额外空间）
*/
public static class Node {
public int value;
public Node next;
public Node(int data) {
this.value = data;
}
}
public static Node listPartition1(Node head, int pivot) {
if (head == null) {
return head;
}
Node cur = head;
int i = 0;
//用于查看共有几个节点
while (cur != null) {
i++;
cur = cur.next;
}
Node[] nodeArr = new Node[i];
i = 0;
cur = head;
//将链表里的数遍历到数组中
for (i = 0; i != nodeArr.length; i++) {
nodeArr[i] = cur;
cur = cur.next;
}
/**荷兰国旗问题：小的放左边，等于的放中间，大于的放右边
* nodeArr：待划分的数组
* pivot：划分值
*/
arrPartition(nodeArr, pivot);
//再将数组中的数串起来成为一个链表
for (i = 1; i != nodeArr.length; i++) {
nodeArr[i - 1].next = nodeArr[i];
}
nodeArr[i - 1].next = null;
//因为调整的时候可能把链表头也顺带替换，所以需要返回链表头
return nodeArr[0];
}
public static void arrPartition(Node[] nodeArr, int pivot) {
int small = -1;
int big = nodeArr.length;
int index = 0;
while (index != big) {
/**
* 精简的荷兰国旗问题：把小于pivot放左边，等于pivot放中间，大于pivot的放右边
*/
if (nodeArr[index].value < pivot) {
swap(nodeArr, ++small, index++);
} else if (nodeArr[index].value == pivot) {
index++;
} else {
swap(nodeArr, --big, index);
}
}
}
public static void swap(Node[] nodeArr, int a, int b) {
Node tmp = nodeArr[a];
nodeArr[a] = nodeArr[b];
nodeArr[b] = tmp;
}
/**
* 算法思想：设立三个区链表，大于等于小于：每个数进行比较，小于划分值放小于区链表，等于划分值放等于区链表......
* 当小于区进来一个数了（每次进来那个数，就将原来里面链表那个数与原链表断开（可以理解为删除）），那么链头和链尾都指向这个数字，此时如果再进来一个数，就将这两个数连起来，链尾指向第二个数，以此类推.大于等于与此相同。
* 直到小于等于大于区链表划分完毕，再将小于区的链尾指向等于区的链头，将等于区的链尾指向大于区的链头，这样就将三个链表连了起来，就达成了目的。
*/
public static Node listPartition2(Node head, int pivot) {
Node sH = null; // small head 小于区的头
Node sT = null; // small tail 小于区的尾
Node eH = null; // equal head 等于区的头
Node eT = null; // equal tail 等于区的尾
Node mH = null; // big head 大于区的头
Node mT = null; // big tail 大于区的尾
Node next = null; // save next node 这里用来存储下一个节点
// every node distributed to three lists
while (head != null) {
//要用next记录下个节点的地址
next = head.next;
//不然这里就成空了
head.next = null;
if (head.value < pivot) {
if (sH == null) {
//如果为null，代表第一次往小于区链表添加节点，令头和尾都指向这个节点
sH = head;
sT = head;
} else {
//这个listPartition2方法返回的是next作为原链表的头，每次执行while语句都是一个新的头
sT.next = head;
sT = head;
}
} else if (head.value == pivot) {
if (eH == null) {
eH = head;
eT = head;
} else {
eT.next = head;
eT = head;
}
} else {
if (mH == null) {
mH = head;
mT = head;
} else {
mT.next = head;
mT = head;
}
}
head = next;
}
/**
* small and equal reconnect：
* 小于区尾巴 与等于区链表尾 连接起来，
* 等于区尾巴和大于区头连接起来，
* 三个链表形成一个新链表
*/
// 如果有小于区域
if (sT != null) {
sT.next = eH;
eT = eT == null ? sT : eT; // 下一步，谁去连大于区域的头，谁就变成eT
}
// 上面的if，不管跑了没有，et
// all reconnect
if (eT != null) { // 如果小于区域和等于区域，不是都没有
eT.next = mH;
}
/**
* 如果小于区的头不为null，则代表有小于区，则把小于区的头返回
* 如果为null，则去看等于区的东西，如果还是为null，则直接返回大于区的头（代码刚开始已经判断链表不为null了）
*/
return sH != null ? sH : (eH != null ? eH : mH);
}
public static void printLinkedList(Node node) {
System.out.print("Linked List: ");
while (node != null) {
System.out.print(node.value + " ");
node = node.next;
}
System.out.println();
}
public static void main(String[] args) {
Node head1 = new Node(7);
head1.next = new Node(9);
head1.next.next = new Node(1);
head1.next.next.next = new Node(8);
head1.next.next.next.next = new Node(5);
head1.next.next.next.next.next = new Node(2);
head1.next.next.next.next.next.next = new Node(5);
printLinkedList(head1);
// head1 = listPartition1(head1, 4);
head1 = listPartition2(head1, 5);
printLinkedList(head1);
}
}
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给定两个可能有环也可能无环的单链表，头节点head1和head2。请实现一个函数，如果两个链表相交，请返回相交的第一个节点。如果不相交，返回null。
【要求】
如果两个链表长度之和为N，时间复杂度请达到O（N），额外空间复杂度请达到O（1）。
public class FindFirstIntersectNode {
public static class Node {
public int value;
public Node next;
public Node(int data) {
this.value = data;
}
}
public static Node getIntersectNode(Node head1, Node head2) {
if (head1 == null || head2 == null) {
return null;
}
Node loop1 = getLoopNode(head1);
Node loop2 = getLoopNode(head2);
if (loop1 == null && loop2 == null) {
return noLoop(head1, head2);
}
if (loop1 != null && loop2 != null) {
return bothLoop(head1, loop1, head2, loop2);
}
return null;
}
// 找到链表第一个入环节点，如果无环，返回null
public static Node getLoopNode(Node head) {
if (head == null || head.next == null || head.next.next == null) {
return null;
}
// n1 慢 n2 快
Node n1 = head.next; // n1 -> slow
Node n2 = head.next.next; // n2 -> fast
while (n1 != n2) {
if (n2.next == null || n2.next.next == null) {
return null;
}
n2 = n2.next.next;
n1 = n1.next;
}
n2 = head; // n2 -> walk again from head
while (n1 != n2) {
n1 = n1.next;
n2 = n2.next;
}
return n1;
}
// 如果两个链表都无环，返回第一个相交节点，如果不想交，返回null
public static Node noLoop(Node head1, Node head2) {
if (head1 == null || head2 == null) {
return null;
}
Node cur1 = head1;
Node cur2 = head2;
int n = 0;
while (cur1.next != null) {
n++;
cur1 = cur1.next;
}
while (cur2.next != null) {
n--;
cur2 = cur2.next;
}
if (cur1 != cur2) {
return null;
}
// n : 链表1长度减去链表2长度的值
cur1 = n > 0 ? head1 : head2; // 谁长，谁的头变成cur1
cur2 = cur1 == head1 ? head2 : head1; // 谁短，谁的头变成cur2
n = Math.abs(n);
while (n != 0) {
n--;
cur1 = cur1.next;
}
while (cur1 != cur2) {
cur1 = cur1.next;
cur2 = cur2.next;
}
return cur1;
}
// 两个有环链表，返回第一个相交节点，如果不想交返回null
public static Node bothLoop(Node head1, Node loop1, Node head2, Node loop2) {
Node cur1 = null;
Node cur2 = null;
if (loop1 == loop2) {
cur1 = head1;
cur2 = head2;
int n = 0;
while (cur1 != loop1) {
n++;
cur1 = cur1.next;
}
while (cur2 != loop2) {
n--;
cur2 = cur2.next;
}
cur1 = n > 0 ? head1 : head2;
cur2 = cur1 == head1 ? head2 : head1;
n = Math.abs(n);
while (n != 0) {
n--;
cur1 = cur1.next;
}
while (cur1 != cur2) {
cur1 = cur1.next;
cur2 = cur2.next;
}
return cur1;
} else {
cur1 = loop1.next;
while (cur1 != loop1) {
if (cur1 == loop2) {
return loop1;
}
cur1 = cur1.next;
}
return null;
}
}
public static void main(String[] args) {
// 1->2->3->4->5->6->7->null
Node head1 = new Node(1);
head1.next = new Node(2);
head1.next.next = new Node(3);
head1.next.next.next = new Node(4);
head1.next.next.next.next = new Node(5);
head1.next.next.next.next.next = new Node(6);
head1.next.next.next.next.next.next = new Node(7);
// 0->9->8->6->7->null
Node head2 = new Node(0);
head2.next = new Node(9);
head2.next.next = new Node(8);
head2.next.next.next = head1.next.next.next.next.next; // 8->6
System.out.println(getIntersectNode(head1, head2).value);
// 1->2->3->4->5->6->7->4...
head1 = new Node(1);
head1.next = new Node(2);
head1.next.next = new Node(3);
head1.next.next.next = new Node(4);
head1.next.next.next.next = new Node(5);
head1.next.next.next.next.next = new Node(6);
head1.next.next.next.next.next.next = new Node(7);
head1.next.next.next.next.next.next = head1.next.next.next; // 7->4
// 0->9->8->2...
head2 = new Node(0);
head2.next = new Node(9);
head2.next.next = new Node(8);
head2.next.next.next = head1.next; // 8->2
System.out.println(getIntersectNode(head1, head2).value);
// 0->9->8->6->4->5->6..
head2 = new Node(0);
head2.next = new Node(9);
head2.next.next = new Node(8);
head2.next.next.next = head1.next.next.next.next.next; // 8->6
System.out.println(getIntersectNode(head1, head2).value);
}
}